$ g(x) = \int_{0}^x\sqrt{10t + 1}\,dt\,$ $ g'(8)\, =$
The Fundamental Theorem of Calculus If $~ g(x)=\int_a^xf(t)\,dt\,$, then $~g^\prime (x)=f(x)\,$ This only works if $f$ is continuous on $[a,b]$. Thankfully, the function $f(t) = \sqrt{10t + 1}$ is continuous on $[0,8]$. Applying the theorem We're given: $ g(x) = \int_{0}^x\sqrt{10t + 1}\,dt\,$ So the theorem tells us: $ g\,^\prime(x) =\sqrt{10x + 1}$ Evaluating $g'(8)$ $ g'(8) = \sqrt{10(8) + 1}=\sqrt{81}=9$ The answer: $g'(8)=9$